Answer: Belongs to the group 2A
Explanation:
As you can see, the first two ionization energies are close and low, meaning that this element ionizates easily.
Not only loses easily the first electron, but the second too
To remove the third electron you requiered a huge amount of energy
Now, elements easily ionizable are the ones from group IA, group 2A and transition metals.
The last ones have mixed characteristics in matter of how many electrons you can remove from them, so they are not a family.
Now the question: group I or group II ?
The elements of group I have low ionization energies for the first electron but high energies for the second ones.
Being all that said, the unknown element belongs to the Group 2A
As,
Water has a pkw=14
so it can be represented as,
[H+] [OH-] = 1*10^-14
If [H+] = 3*10^-5M
[OH-] = (1*10^-14) / ( 3*10^-5)
[OH-] = 3.3*10^-9 M
Using Gay-Lussac's Law, pressure is proportional to (absolute) temperature in Kelvin. We first convert the temperature values to Kelvin: 110 C = 383.15 K, while 65 C = 338.15 K.
P1/T1 = P2/T2
22.5/383.15 = P2/338.15
P2 = 19.9 psi
Answer:
<h3>The answer is 1.99 g/mL</h3>
Explanation:
The density of a substance can be found by using the formula
From the question
mass = 16.93 g
volume = final volume of water - initial volume of water
volume = 19.7 - 11.2 = 8.5 mL
We have
We have the final answer as
<h3>1.99 g/mL</h3>
Hope this helps you
